Perpendicular components of motion

[leroyjenkens]

I was doing some practice problems on physics classroom website and this is one I think is wrong.

A plane flies northwest out of O'Hare Airport in Chicago at a speed of 400 km/hr in a direction of 150 degrees (i.e., 30 degrees north of west). The Canadian border is located a distance of 1500 km due north of Chicago. The plane will cross into Canada after approximately ____ hours.

And the answer they gave is...

The plane has both a northward and a westward motion. The northward motion towards the Canadian border is dependent upon the component of velocity in the northern direction. To solve this problem, the northern component of the plane's velocity must first be determined.

vNorth = 400 • sine (150 degrees) = 200 km/hr
Once determined, the d = v • t equation can be used to determine the time that it takes the plane to reach the Canadian border.

dNorth = vNorth • t t = dNorth / vNorth = (1500 km) / (200 km/hr) = 7.5 hr


But wouldn't the component of velocity that the sine function calculated be the westward motion? Sine is opposite over hypotenuse, so the opposite would be the westward motion. The northward motion would be the adjacent, right? I'm probably wrong about the answer, but I don't know why they got the answer they did.

Thanks.

 

[rs54]

I have to side with the given answer. When I imagine 30 degrees north of west, I think of facing west and then turn 30 degrees towards the north. So if I make a triangle out of the picture, I’m at the 30 degree angle, the adjacent side is pointing out west from me, the hypotenuse is pointing out in the sort of NW direction I’m facing after I turn. And the opposite is pointing north.

This is also consistent with 150 degrees for an angle in standard position where +x, or east is 0 degrees.

 

[leroyjenkens]

I have to side with the given answer. When I imagine 30 degrees north of west, I think of facing west and then turn 30 degrees towards the north. So if I make a triangle out of the picture, I’m at the 30 degree angle, the adjacent side is pointing out west from me, the hypotenuse is pointing out in the sort of NW direction I’m facing after I turn. And the opposite is pointing north.

This is also consistent with 150 degrees for an angle in standard position where +x, or east is 0 degrees.

Thanks for the response. I was going by the picture they gave, which would put the adjacent as the vertical line...

Using that picture, is their answer wrong, or do they still get the right answer using it?

 

[rs54]

I think the picture looks OK.

The vertical red line is just showing how far north of Chicago the border is.

The slanted red line is the velocity vector of the plane and they didn’t draw any triangle showing its components. The y-component of that vector shows the northward speed of the plane and the triangle has to be drawn for that vector.

So the plane is headed at the slanted angle. The 30° angle is how much north it’s turned from a westward direction (so the velocity is 30° north of west, as they said).

See how it is going north less than it is going west? The sine of 30° is smaller than the cosine so if you got confused that would also clue you in that the north should be the sine. And the north component is certainly the opposite side if you draw the triangle for the vector showing the angle at the starting point of the vector.

The 150° angle is all the way from the east (+x, the standard starting point for an angle) to the vector. Its sign is also 1/2. Some people find it less confusing to use the angle from the standard starting point. Then the sine will be the y-component and the cosine will be the x-component (in this case the x-component is negative west being the –x direction). If you do it that way you have to be good at the trig for angles all the way around the unit circle.

http://12784015027620697233-a-g.googlegroups.com/web/Plane%20from%20Chicago.gif?gsc=yd_LaAsAAAB01JrUNAWIL3lBNWDabkJb

 

[leroyjenkens]

What confused me is how the angles are positioned. It has the vertical line and the slanted line almost forming a right triangle. So what I did was added a line at the top along that dotted line, connecting the two, which is what I thought they did. If I kept the triangle like that, the opposite side would be horizontal and the adjacent side is the vector with the 1500 next to it.
Thanks.